Question

An object with initial temperature 130 ∘ F is submerged in large tank of water whose temperature is 50 ∘ F . Find a formula for F ( t ) , the temperature of the object after t minutes, if the cooling constant is k = − 0.2 . Remember Newton's Law of Cooling (the rate of change of temperature with respect to time is equal to k times the difference between the temperature of the object and the surrounding temperature) ! :)

Answer 1

Answer:

T = 50 + 80e^(-0.2t)

Explanation:

Newton's law of cooling says the rate of change of temperature with respect to time is proportional to the temperature difference:

dT/dt = k (T − Tₐ)

Separating the variables and integrating:

dT / (T − Tₐ) = k dt

ln (T − Tₐ) = kt + C

T − Tₐ = Ce^(kt)

T = Tₐ + Ce^(kt)

Given that Tₐ = 50 and k = -0.2:

T = 50 + Ce^(-0.2t)

At t = 0, T = 130.

130 = 50 + Ce^(0)

130 = 50 + C

C = 80

Therefore:

T = 50 + 80e^(-0.2t)

Answer 2

The formula for the object cooling is $T(t) = 50 + 80\cdot e^{-0.2\cdot t}$, where $t$ is in minutes.

The object is cooled by heat mechanism of Convection, Convection is a Heat Transfer mechanism in which is a solid object is cooled due to a fluid in motion and is described by the Newton's Law of Cooling, whose Differential Equation is:

$\frac{dT}{dt} = -r\cdot (T-T_{m})$ (1)

Where:

$T$ - Temperature of the solid, in degrees Fahrenheit.

$r$ - Cooling rate, in $\frac{1}{min}$.

$T_{m}$ - Water temperature, in degrees Fahrenheit.

The solution of this Differential Equation is:

$T(t) = T_{m} + (T_{o}-T_{m})\cdot e^{-r\cdot t}$ (2)

Where $T_{o}$ is the initial temperature of the solid, in degrees Fahrenheit.

If we know that $T_{m} = 50\,^{\circ}F$, $T_{o} = 130\,^{\circ}F$ and $r = 0.2$, then the formula for the object cooling is:

$T(t) = 50 + 80\cdot e^{-0.2\cdot t}$

The formula for the object cooling is $T(t) = 50 + 80\cdot e^{-0.2\cdot t}$, where $t$ is in minutes.

Here is a question related to the Newton's Law of Cooling: https://brainly.com/question/13724658